Poisson's Ratio - Longitudinal Strain and Lateral Strain

What is Poisson’s Ratio?

Imagine a piece of rubber, in the usual shape of a cuboid. Then imagine pulling it along the sides. What happens now?

Poisson's ratio

It will compress in the middle. If the original length and breadth of the rubber are taken as L and B respectively, then when pulled longitudinally, it tends to get compressed laterally. In simple words, length has increased by an amount dL and the breadth has increased by an amount dB.

In this case,

εt =  \( \frac {-dB}{B}\)

εl= \( \frac {dL}{L}\)

The formula for Poisson’s ratio is,

μ = \( \frac {-ε_t}{ε_l}\)

where,

εt is the Lateral or Transverse Strain

εl is the Longitudinal or Axial Strain

μ is the Poisson’s Ratio

Strain on its own is defined as the change in dimension (length, breadth, area…) divided by the original dimension.

Poisson Effect: When a material is stretched in one direction, it tends to compress in the direction perpendicular to that of force application and vice versa. The measure of this phenomenon is given in terms of Poisson’s ratio. For example, a rubber band tends to become thinner when stretched.

Definition of Poisson’s ratio for a material:

It is the ratio of transverse contraction strain to longitudinal extension strain, in the direction of stretching force. There can be a stress and strain relation which is generated with the application of force on a body. For tensile deformation, Poisson’s ratio is positive and for compressive deformation it is negative, as in, negative Poisson ratio suggests that the material will exhibit a positive strain in the transverse direction, even though the longitudinal strain is positive as well. For most materials, the value of Poisson’s ratio lies in the range, 0 to 0.5.

A few examples of Poisson ratio is given below for different materials.

Rubber = 0.49

Aluminium = 0.32

Concrete = 0.2

Cork = 0

Consider the following situation. You know that the value of Poisson’s ratio for concrete is 0.2 but if the concrete cracks, the value is 0! Why might that be?


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Practise This Question

If the flow of water in the tank stops, what will be the depth of water in the tank. Assume that the level of oil outside the tank does not increase.